Question

We are given the following atomic masses:
`._92Pu^(238)=238.05079u,`
`._90Th^(234)=234.04363u`,
`._91Pa^(237)=237.05121, ._1H^1=1.00783`,
`._2He^2=4.00260u`
(a) Calculate the energy released during `alpha` decay of `._92U^(238)`,
(b) Calculate the kinetic energy of emitted `alpha` particles,
(c) show that `._94Pu^(238)` cannot spontaneously emit a proton.

Answer 1

(a) `._92U^(238) to ._90Th^(234)+._2He^4`
`Deltam=(238.05079-234.04363-4.00260)`
`=0.00456u`
Energy released, `Q=0.00456xx931.5MeV`
`=4.25MeV`
(b) K.E. of `alpha` particle =`((A-4)/A)xxQ`
`=(238-4)/238xx4.25MeV`
`=4.18MeV`
(c) If `._92U^(238)` emits a proton spontaneously,
`._92U^(238) to ._91Pa^(237)+._1H^1`
`Deltam=(238.05079-237.05121-1.00783) u`
`:. Q=-0.00825u=-0.00825xx931.5MeV`
`=-7.68MeV`
As the Q value is negative, the process cannot proceed spontaneously.